Please answer this passionately and with full detail! I need the hel.pdf

•

0 likes•2 views

Please answer this passionately and with full detail! I need the help big time. Question: There are pathogenic strains of E coli that attach to epithelial cells in the bladder and cause disease there. Some of these strains possess three different kinds of protein appendages: fimbriae, conjugation pili, and flagella. What are the functions of each of these appendages? Which is the most important for infecting the bladder? Why do you think that? Which is least important for infecting the bladder? Why do you think that? Solution Although E.coli reside in the gut as normal flora, but some strains of this microbe can be infectious in nature. An example of this strain is E. coli O157:H7. This strain of E. coli contains infectious elements which can be easily transmitted to host and cause disease. One of the most common type of infection occuring due to E. coli is renal and urinary tract infection. In this kind of infection, the E. coli releases toxins in the body of the host and causes disease. Some of the infectious strains possess different kinds of appendages like fimbriae, pili and flagella. Their unique functions are discussed as below: Fimbriae: Fimbriae are appendages which help in attachment of the bacteria on the surface of attachment. This helps the bacteria to reside and colonize at a place and undergo infectious cycle. Fimbriae can be found all over the surface. Conjugation pili: These are hollow membranous extensions which help in transfer of genetic material from one microbe to another. The pili help in transfer of plasmids and thus bring about sexual reproduction in bacteria. Flagella: Flagella represents the appendage of motion and movement in bacteria. A typical movement of flagella is important to modulate forward, reverse or tangling motion of the bacteria. Which is the most important for infecting the bladder? As described above, it is most important for a bacteria to attach to the surface and colonize over there to cause any infection. It should be noted here that a few bacteria are not sufficient enough to cause an infection. A large colony is required to induce a diseased state. For that to be accomplished, the bacteria must attach to the surface firmly. This can be brought about only if they have fimbriae. Thus, fimbriae are most important for infection to occur. Which is least important for infecting the bladder? As it can be observed from the above explanation, pili are the structural appendages required for transmission of genetic material from one bacteria to another. These are also responsible for generation of virulence in non-virulent strains by transmission of virulent plasmid. However, they are not directly related with causing infection in bladder since their presene is meant for transfer of information among bacteria only..

Education

generation of virulence in non-virulent strains by transmission of virulent plasmid. However,
they are not directly related with causing infection in bladder since their presene is meant for
transfer of information among bacteria only.

classify domian of life differencesSolutionClassify the domain.pdf

arihantmum

Compare and contrast the development of a WBS in traditional project management versus development in an agile environment. Solution An Agile based WBS is organized around units of end user functionality - commonly referred to as user stories. Each user story represents the work to deliver that increment of functionality. In Agile based WBS, Features are decomposed into Epics (large user stories) and Epics into User Stories. User stories are decomposed to represent functionality that can be completed within one iteration - typically one to four weeks depending on the team/project. Each leaf level WBS element represent the working software to be produced through analysis, design, coding and testing of the requirement in the user story. Because each user story is atomic, stories can be re-prioritized, added and or removed without changing the scope, provided that the total size/effort of stories within the scope of the project does not change. Additionally, iterative development facilitates multiple releases (deployments) within a project. So, user stories are typically scheduled within an iteration and a release. Note, that until an iteration has started, the stories within that iteration can be changed, as part of iteration planning. In case of traditional project management WBS is organized around the primary products of the project (or planned outcomes) instead of the work needed to produce the products (planned actions). Since the planned outcomes are the desired ends of the project, they form a relatively stable set of categories in which the costs of the planned actions needed to achieve them can be collected. A well-designed WBS makes it easy to assign each project activity to one and only one terminal element of the WBS. In addition to its function in cost accounting, the WBS also helps map requirements from one level of system specification to another, for example a requirements cross reference matrix mapping functional requirements to high level or low level design documents..

Compare and contrast the development of a WBS in traditional project.pdf

arihantmum

B. You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement—\"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond.? Solution You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement— \"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. Chance: It plays a part in evolution (for example, random mutations in a gene pool can give rise to few new traits), but evolution does not depend ONLY on chance in creating new organisms, proteins or any other entities. This is quite opposite to natural selection. Natural selection is the principal which is known for the mechanism of evolution. By chance some mutation caused in the gene pool and it gave to new acceptable traits. If they found adjustable to the environment, they continues as a new species. The \"undesirable\" (nonadaptive) ones will be thus elimated from that gene pool. As long as the forces of selection stay constant, natural selection pushes the evolution in only one direction and produce new desired structures. Thus chance plays a part in evolution. Nature plays a major role in this. Fossil studies support this theory/ process. Where we can see many intermediate or failed species in between major species. Archaeopteryx, a link between reptiles and birds, extinctinted soon is a good example of intermediate species. Haldane and Oparin proposed a theory of the origin of life on Earth through the transformation, during the gradual chemical evolution of particles that contain carbon in the primordial soup shows an example of Evolution. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond? As a creation scientist i do not agree with science and evolution part, all this is created by God and thus it follows biblical roots. All that it is in the nature is creation of God, all of a sudden it appeared on the earth. According to us GOD created all species like birds,animals etc., later he wanted to create a genius brained animal, so he created humans. And God creation according to us all this is due to CHOICE not because of CHANCE.Thus it involved logical and systematic process..

B. You are an evolutionary biologist in a heated argument with a cre.pdf

arihantmum

Assume Hashtable is a simple array of size 8, with indices 0..7. Numeric keys are mapped by a Hashfunction that gives the mod(8,n) value for any key “n” yielding the Hashtable index for that key (0..7). A Hashtable entry is null unless a key exists with that index as its hashed index; if so, the Hashtable entry points to the first node of a linked list of keys with that hash index. The last node on this linked list has a null reference for the next referenced node. Assume the occurrence of a linked list node is represented by the object “Node” and its “Data” and “NextRef” attributes. Create pseudocode for the \"Remove\" operation. Solution Removal of a Node from a Singly Linked List To complete deletion of firstNode in the list we have to change Head pointing to NextRef of firstNode. Pseudocode: firstNode = Head Head = firstNode->NextRef free firstNode Complexity: Time Complexity: O(1) Space Complexity: O(1) Sourcecode: int delNodeData(int num) { struct Node *prev_ptr, *node; node=Head; while(cur_ptr != NULL) { if(cur_ptr->Data == num) { if(cur_ptr==Head) { Head=cur_ptr->NextRef; free(node); return 0; } else { prev_ptr->NextRef=node->NextRef; free(node); return 0; } } else { prev_ptr=node; node=node->NextRef; } } printf(\"\ Element %d is not found in the List\", num); return 1; } Deleting Last Node in the Singly Linked List Traverse to Last Node in the List using two pointers namely prevNode and node. Once node reaches the last Node in the list point NextRef in prevNode to NULL and free the node. Pseudocode: node = head forever: if node->NextRef == NULL break prevNode = node node = node->NextRef prevNode->NextRef = NULL free Node Complexity: Time Complexity: O(n) Space Complexity: O(1) Deleting Node from position \'p\' in the List To delete a Node at the position \'p\' we have to first traverse the list until we reach the position \'p\'. For this case have to maintain two pointers namely prevNode and Node. Since Singly Linked Lists are uni-directional we have to maintain the information about previous Node in prevNode. Once we reach the position \'p\' we have to modify prevNode NextRef pointing to Node NextRef and free Node. Pseudocode: curNode = head curPos = 1 forever: if curPos == P || Node == NULL break prevNode = node node = node->NextRef curPos++ if noode != NULL: prevNode->NextRef = node->NextRef free curNode Complexity: Time Complexity: O(n) worst case Space Complexity: O(3) Sourcecode: int delNodeLoc(int loc) { struct Node *prev_ptr, *node; int i; node=Head; if(loc > (length()) || loc <= 0) { printf(\"\ Deletion of Node at given location is not possible\ \"); } else { // If the location is starting of the list if (loc == 1) { Head=node->NextRef; free(node); return 0; } else { for(i=1;iNextRef; } prev_ptr->NextRef=node->NextRef; free(node); } } return 1; }.

Assume Hashtable is a simple array of size 8, with indices 0..7. Num.pdf

arihantmum

Amon the following, which has the lowest levels of dissolved iron? a. Atlantic Ocean b. Indian Ocean c. Pacific Ocean d. The temperate regions of all oceans Solution The correct option is Pacific ocean because of the low dust sedimentation and iron is readily soluble in low oxygen regions and the continental margins. It was believed that iron remained in the localized areas but didn\'t contribute to the overall iron content of the ocean. As it is explained by Nielsen, as soon as the iron-rich fluids hit the seawater with high oxygen content, the iron gets dumped out and is not contributed to the overall amount of the dissolved iron..

Amon the following, which has the lowest levels of dissolved iron.pdf

arihantmum

A box contains 10 red balls and 40 black balls. Two balls are drawn .pdf

arihantmum

As late as 1992, the United States was running budget deficits of nearly $300 billion. During the remainder of the 1990\'s, deficits declined and became surpluses. As the new century began, these surpluses again turned into deficits. •Explain the decline in deficits and subsequent surpluses in the late 1990\'s. •Explain the return to deficit spending since the turn of the century. •Consider the causes of the deficits and surpluses and provide your own insight as to whether these surpluses or deficits have a \"positive\" or \"negative\" effect on our economy Solution As the 1990s came to an end, virtually all financial-market observers and politicians rejoiced that budget deficits were gone and that the U.S. had entered a period of fiscal stability. Politicians on both sides of the aisle viewed the new world of budget surpluses as an opportunity to save Social Security, to cut Federal debt, and to lower taxes. Budget deficits or budget surpluses are neither good nor bad, they are fiscal policy options that can stabilize or de-stabilize an economy. The U.S. economy, under Reagan, Bush, and early Clinton, experienced higher not lower rates of economic growth at a time when budget deficits were at record levels and tax rates were on the decline. Over the past 20 years a period of record budget deficits the stock market increased from 950 to over 11,000, providing ample evidence that budget deficits are not necessarily bad for an economy.On the other hand, budget surpluses are not necessarily good. As the budget surpluses began to expand dramatically in late 1999 and into 2000, the stock market appeared to be reversing the record up-trend of the 1980s and \'90s. One possible reason for the reversal: the growing budget surplus was sapping the savings of the private sector. Consumers were able to maintain their spending only through expanding debt to record level. In the late 90s, the confluence of strong economic growth, decelerating federal government spending, and substantial increases in tax rates (revenues) had the unexpected effect of producing enormous budget surpluses. As unexpected budget surpluses burst on the scene in the late 1990s, most mainstream economists weren’t sensitive to the contractionary impact that surpluses can have on the economy. The reason was simple that none of their economic models ever included the forecast of a substantial surplus. Ronald Reagan’s was the vice president of the USA for eight years and knew firsthand how difficult it was to bring the federal budget deficit down from 6 percent of GDP in 1983 to 2.8 percent in Reagan’s last budget. Bush knew that the many budget deals of the 1980s had been hard fought and always involved higher taxes as part of the deficit reduction, Reagan signed into law 11 major tax increases that raised taxes by $133 billion in 1988 or 2.7 percent of GDP.Bush administration economists believed that an easier monetary policy was the key to stimulating growth. But Federal Reserve Board.

As late as 1992, the United States was running budget deficits of ne.pdf

arihantmum

A bacteriophage population is introduced to a bacterial colony that is in stationary phase with rapidly depleting nutrients. Appraise the following scenarios and select which one would favor long-term survival of the bacteriophage. A. The bacteriophages infect the remaining viable cells and integrate their genetic material into the host cell genome thereby preserving their genome. B. The bacteriophages do not infect the bacterial cells since they are dying; they remain inert outside of the cells. C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment. Solution Answer : C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment..

A bacteriophage population is introduced to a bacterial colony that .pdf

arihantmum

Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee shop independently. Xavier\'s arrival time is X and Yolanda\'s arrival time is Y, where X and Y are measured in minutes after noon. The individual density functions are given. (Xavier arrives sometime after noon and is more likely to arrive promptly than late. Yolanda always arrives by 12:10 PM and is more likely to arrive late than promptly.) After Yolanda arrives, she\'ll wait for up to 20 minutes for Xavier, but he won\'t wait for her. Find the probability that they meet. (Round your answer to three decimal places.) f1(x) = {e^-x if x gt o and o if x Solution.

Xavier and Yolanda both have classes that end at noon and they agree .pdf

arihantmum

1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia (15pts). a. What are the parental gametes? (2pts) b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible genotypes. (2 pts) c. What is the gamete frequency if the two genotypes in (b) were crossed with each other? (4pts)? (4pts) d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts) Solution Both the conditions are recessive. Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for beta thalessemia. Genotype is LLbb Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta- thalessemia. Genotype is llBb The possible gametes produced by these parents: Nikoleta: Lb Husband: lB, lb Possible genotype of the F1 offspring: LlBb or Llbb Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype should be Llbb. Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb). Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2 The following is the phenotype frequency: LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3 LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3 llBb (Low platelet count and no beta thallasemia) = 1 llbb (Low platelet count and beta thallasemia) = 1 Thus phenotypic ratio stands as: Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1.

1. Low platelet count is a recessively inherited trait. Reevaluation.pdf

arihantmum

More from arihantmum (20)

In 2012, the percent of American adults who owned cell phones and us.pdf

In a multiple regression model, the error term e is assumed tohave.pdf

I need help with this one method in java. Here are the guidelines. O.pdf

help me Java projectI put problem and my own code in the linkmy .pdf

Heading ibe the following picture. Main computer UART DTE Serial chan.pdf

Explain how TWO (2) different structural features can be used to dis.pdf

Explain two (2) alternative ways in which plants can obtain nutrient.pdf

Couldnt find the right subject that it belongs to...There ar.pdf

Consider the following C code snippet C codevoid setArray(int.pdf

Describe a mechanism that the body uses to prevent a mutation from be.pdf